從3個表中選擇數據

Question

我想從數據庫中獲取一些數據。

1. username從接觸表

登錄表

email在兩個表tutor和institute

與2個值檢查這是到目前爲止我的代碼：

SELECT s. * , c.email, l.username 
FROM (
     SELECT contact_id AS id, 
        login_id, 
        username, 
        tutor_code AS code, 
        tutor_name AS Name, 
        'tutor' AS profile 
     FROM tutors 
     WHERE tutor_code = $code AND tutor_name = '$name' 
     UNION ALL 
     SELECT contact_id AS id, 
        login_id, 
        username, 
        institute_code AS code, 
        institute_name AS Name, 
        'institute' AS profile 
     FROM institutes 
     WHERE institute_code = $code AND institute_name = '$name' 
     ) 
INNER JOIN contact c ON s.id = c.contact_id 
INNER JOIN login l ON s.login_id = l.login_id 

此查詢不起作用，並且有錯誤消息。

1054 - 在 '字段列表'

UPDATE

SELECT s. * , c.email, l.username 
FROM (
     SELECT contact_id AS id, 
        login_id, 
        username, 
        tutor_code AS code, 
        tutor_name AS Name, 
        'tutor' AS profile 
     FROM tutors 
     WHERE tutor_code = $code AND tutor_name = '$name' 
     UNION ALL 
     SELECT contact_id AS id, 
        login_id, 
        username, 
        institute_code AS code, 
        institute_name AS Name, 
        'institute' AS profile 
     FROM institutes 
     WHERE institute_code = $code AND institute_name = '$name' 
     )s 
INNER JOIN contact c ON s.id = c.contact_id 
INNER JOIN login l ON s.login_id = l.login_id 

Answer 1

因爲它似乎要檢索您的username從login，列username最有可能不tutors和/或institutes存在，也沒有必要，因爲你是在login_id加盟加盟login ，我想你可以只從子查詢中刪除username列：

SELECT s. * , c.email, l.username 
FROM (
     SELECT contact_id AS id, 
        login_id, 
        --username, 
        tutor_code AS code, 
        tutor_name AS Name, 
        'tutor' AS profile 
     FROM tutors 
     WHERE tutor_code = $code AND tutor_name = '$name' 
     UNION ALL 
     SELECT contact_id AS id, 
        login_id, 
        --username, 
        institute_code AS code, 
        institute_name AS Name, 
        'institute' AS profile 
     FROM institutes 
     WHERE institute_code = $code AND institute_name = '$name' 
     ) s 
INNER JOIN contact c ON s.id = c.contact_id 
INNER JOIN login l ON s.login_id = l.login_id 

我還添加別名s您subuqery因爲我以爲我T的遺漏是一個錯字，因爲它會在沒有拋出一個語法錯誤

Answer 2

這是在子查詢中username場未發現未知列 '用戶名'，你必須在子查詢中包含login表。

SELECT s. * , c.email, l.username 
FROM (
     SELECT contact_id AS id, 
        login_id, 
        l1.username, 
        tutor_code AS code, 
        tutor_name AS Name, 
        'tutor' AS profile 
     FROM tutors t 
     INNER JOIN login l1 ON l1.login_id = t.login_id 
     WHERE tutor_code = $code AND tutor_name = '$name' 
     UNION ALL 
     SELECT contact_id AS id, 
        login_id, 
        l2.username, 
        institute_code AS code, 
        institute_name AS Name, 
        'institute' AS profile 
     FROM institutes i 
     INNER JOIN login l2 ON l2.login_id = i.login_id 
     WHERE institute_code = $code AND institute_name = '$name' 
     ) 
INNER JOIN contact c ON s.id = c.contact_id 
INNER JOIN login l ON s.login_id = l.login_id 

Answer 3

沒有必要從tutors和institutes表調用username和使用as abc當你的子查詢關閉像

SELECT s. * , c.email, l.username 
FROM (
     SELECT contact_id AS id, 
        login_id, 
        tutor_code AS code, 
        tutor_name AS Name, 
        'tutor' AS profile 
     FROM tutors 
     WHERE tutor_code = $code AND tutor_name = '$name' 
     UNION ALL 
     SELECT contact_id AS id, 
        login_id, 
        institute_code AS code, 
        institute_name AS Name, 
        'institute' AS profile 
     FROM institutes 
     WHERE institute_code = $code AND institute_name = '$name' 
     ) as s 
INNER JOIN contact c ON s.id = c.contact_id 
INNER JOIN login l ON s.login_id = l.login_id 

這個查詢將返回重複條目，因爲您在union中使用ALL。

希望它適合你。